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Find an infinite power series of the form $\sum_n a_n z^n$ with radius of convergence 1 that converges for every $z$ such that $|z|=1$ except when $z = z_1, z_2, \ldots, z_m$ where $z_1$, $z_2,\ldots, z_m$ are complex numbers with modulus 1.

PseudoNeo
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guest10923
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  • Do you know an example of power series of radius of convergence 1 which converges for all $z$ s.t. $|z| = 1$, except for $z = 1$? – PseudoNeo Mar 09 '15 at 08:53
  • Yep, $(z^n)/n$ Do I have to consider the complex number in exponential form? Or in terms of sin and cos? – guest10923 Mar 09 '15 at 08:59
  • Well, either will do. But it is probably more simple to use it in exponential form. Now, the only thing you have to do is to modify the $\sum \frac{z^n}n$ example a bit to produce the example you're looking for. First step: how to make an example of a power series converging for all complex numbers of modulus 1 except for $z=z_1$? – PseudoNeo Mar 09 '15 at 09:03
  • Multiply the z by the conjugate of z1? so $(z1^*z)^n/n$? – guest10923 Mar 09 '15 at 09:10
  • Yes. So now you have power series $\sum \frac{(\overline{z_i} z)^n}{n}$, each only diverging on the point $z = z_i$. How to combine them so to get a power series diverging on all of these points? – PseudoNeo Mar 09 '15 at 09:20
  • I am not sure, you can't multiply by all the conjugates as still only one point will converge but you also can't have (z1* +z2...+zn)$z^n$ as this doesn't have a radius of convergence equal to 1. – guest10923 Mar 09 '15 at 09:27
  • No, what you have to do is to multiply all these power series. Do you know how to do that? – PseudoNeo Mar 09 '15 at 09:28
  • Umm, I don't think I have come across it no. I guess it isn't as simple as multiplying the sums together?? – guest10923 Mar 09 '15 at 09:32
  • The technical term is Cauchy Product: http://en.wikipedia.org/wiki/Cauchy_product ; basically, it is the same formula as the one you use to multiply two polynomials. If you do that to all your power series, you will get the example you're looking for. – PseudoNeo Mar 09 '15 at 09:36

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